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zmey [24]
2 years ago
9

American households spent an average of about $52 in 2007 on Halloween merchandise such as costumes, decorations and candy. To s

ee if this number had changed, researchers conducted a new survey in 2008 before industry numbers were reported. The survey included 1,500 households and found that average Halloween spending was $58 per household.The sample mean is __________ dollars, while the claimed population mean is _________ dollars.
Mathematics
1 answer:
Grace [21]2 years ago
6 0

Answer:

The sample mean is$ 52

The claimed population mean is $58

Step-by-step explanation:

American households spent an average of about $52 in 2007 on Halloween merchandise such as costumes, decorations and candy. This tells you that the sample mean is$ 52.

1500 households were surveyed in 2008 and found an average spending of $58 0n Halloween tell us that the claim's population mean is $58

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Mr. Torres took his students to the dolphin show. Each row in the stadium had 11 seats. One adult sat at each end of a row, and
Katena32 [7]

Answer:

3

Step-by-step explanation:

The given data shows that only 1 row was in use. We also know that Mr. Torres is an adult. So for the ease, lets use A for Adult and S for student.

The seating must fulfill the requirements that 4 students must be between adults such that it would be "A-S-S-S-S-A"

On a row of 11 seats, this should be the searing arrangements.

A-S-S-S-S-A-S-S-S-S-A ( for 1 row).

Mr. Torres could be any A .

3 0
2 years ago
Find the values for k so that the intersection of x = 2k and 3x + 2y = 12 lies in the first quadrant.
RUDIKE [14]

Answer:

0 < k < 2

Step-by-step explanation:

<u>In the first quadrant both of x and y get positive values, so </u>

  • x > 0 and y > 0
  • x = 2k, k > 0

<u>And replacing x with 2k in the second equation:</u>

  • 3x + 2y = 12
  • 3*2k + 2y = 12
  • 2y= 12 - 6k
  • y = 6 - 3k

<u>Since y > 0:</u>

  • 6 - 3k> 0
  • 2 - k > 0
  • k < 2

<u>Combining both k > 0 and k < 2, we get:</u>

  • 0 < k < 2
3 0
2 years ago
Read 2 more answers
(25 points, please help! Brainliest!) Mrs. Hanger is painting the following picture of an H to hang in the entryway of her home.
disa [49]

Answer:

the new scale is the same length as the original scale

7 0
1 year ago
Of 560 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compared
frutty [35]

Answer:

a) (0.5256,0.5944)  

c) Criticism is invalid

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 560

Proportion of mislabeled = 56%

\hat{p} = 0.56

a) 90% Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64

Putting the values, we get:

0.56\pm 1.64(\sqrt{\dfrac{0.56(1-0.56)}{560}}) = 0.56\pm 0.0344\\\\=(0.5256,0.5944)

b) Interpretation of confidence interval:

We are 90% confident that the true proportion of all seafood in the country that is mislabeled or misidentified is between 0.5256 and 0.5944 that is 52.56% and 59.44%.

c) Validity of criticism

Conditions for validity:

np > 10\\n(1-p)>10

Verification:

560\times 0.56 = 313.6>10\\560(1-0.56) = 246.4>10

Both the conditions are satisfied. This, the criticism is invalid.

8 0
2 years ago
Nathan had an infection, and his doctor wanted him to take penicillin. Because Nathan’s father and paternal grandfather were all
Sergio039 [100]
Let events
A=Nathan has allergy
~A=Nathan does not have allergy
T=Nathan tests positive
~T=Nathan does not test positive

We are given
P(A)=0.75  [ probability that Nathan is allergic ]
P(T|A)=0.98  [probability of testing positive given Nathan is allergic to Penicillin]

We want to calculate probability that Nathan is allergic AND tests positive
P(T n A)

From definition of conditional probability,
P(T|A)=P(T n A)/P(A)
substitute known values,
0.98 = P(T n A) / 0.75
solving for P(T n A)
P(T n A) = 0.75*0.98 = 0.735

Hope this helps!!
3 0
2 years ago
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